# Projectile Motion without Initial velocity

1. Oct 13, 2009

### Disoo

1. The problem statement, all variables and given/known data
A stone is fired from a slingshot at an angle of 65° from the horizontal. The stone strikes the ground 8.0s later at an altitude 30m lower than the height at which it was released.
At what initial velocity was the stone released?

2. Relevant equations
dy = viyt + 1/2gt2
Viy = Vi sin θ

3. The attempt at a solution
dy = viyt + 1/2gt2
dy = Vi sin θt + 1/2gt2
dy + 30m = Vi sin (65)(8.0s) + 1/2(+9.91m/s2)(8.0s)2
dy =7.25 Vi + 283.92 m

I'm stuck after this stage. Currently I don't see a clear solution. How could I approach the problem?

2. Oct 13, 2009

### rl.bhat

Hi Disoo, welcome to PF.
In your equation you have not used the proper sign convention.
For the projectile motion the equation can be written as
y = yo + vi*sinθ*t - 1/2*g*t^2.
Take y = 0 at ground level.

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